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Math Help - n-epsilon relationship for geometric mean limit

  1. #1
    yms
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    n-epsilon relationship for geometric mean limit

    Any idea how to find at least one "n" such that:


    [ ( x1/n + y1/n ) / 2 ]2n - xy < epsilon


    2 <= x <= a, 2 <= y <= a

    for any epsilon?


    Actually, its well-known that:

    lim [ ( x1/n + y1/n ) / 2 ]2n = xy
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  2. #2
    yms
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    Re: n-epsilon relationship for geometric mean limit

    Even for lim n^(1/n) = 1 it is a problem ...

    (1/n) (log n) < log(1 + eps) , n > ?? ...
    Last edited by yms; November 14th 2012 at 05:30 PM. Reason: typo
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